In the circuit shown in figure- V1 and V2 are two voltmeters having resistances 6000 - and 4000 - - E-M-F of battery is 250 Volts- having negligible internal resistance- Two resistances R1 and R2 are 4000 - and 6000 - respectively- then -

When (s) is open I=2505000=120 A I_1=I2=140 A V_1=V_2=I_1R_1=10000×140=125 V Option 3, 4 are correct. When (s) is closed: R=R_1+R_2=10000 ...

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Byju's Answer

Standard VIII

Physics

Electric Power

In the circui...

Question

In the circuit shown in figure, $V_{1} and V_{2}$ are two voltmeters having resistances $6000 Ω and 4000 Ω$ , E.M.F of battery is $250 Volts,$ having negligible internal resistance. Two resistances $R_{1} and R_{2}$ are $4000 Ω$ and $6000 Ω$ respectively, then :

V_{2}

when switch S is close will be

150 V

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V_{1}

when switch S is close will be

100 V

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V_{1}

when switch S is open will be

125 V

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V_{2}

when switch S is open will be

125 V

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Solution

The correct options are
A
$V_{2}$ when switch S is close will be $150 V$
B
$V_{1}$ when switch S is close will be $100 V$
C
$V_{1}$ when switch S is open will be $125 V$
D
$V_{2}$ when switch S is open will be $125 V$

When (s) is open

$I = \frac{250}{5000} = \frac{1}{20} A I_{1} = \frac{I}{2} = \frac{1}{40} A V_{1} = V_{2} = I_{1} R_{1} = 10000 \times \frac{1}{40} = 125 V$

Option 3, 4 are correct.

When (s) is closed:

$R = R_{1} + R_{2} = 10000 I = \frac{250}{10000} = \frac{1}{40} V_{1} = \frac{1}{40} \times 4000 = 100 V [Option 2 correct] V_{2} = \frac{1}{40} \times 6000 = 150 V [Option 1 correct]$

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In the circuit shown in figure, V1 and V2 are two voltmeters having resistances 6000 Ω and 4000 Ω , E.M.F of battery is 250 Volts, having negligible internal resistance. Two resistances R1 and R2 are 4000 Ω and 6000 Ω respectively, then :

In the circuit shown in figure, $V_{1} and V_{2}$ are two voltmeters having resistances $6000 Ω and 4000 Ω$ , E.M.F of battery is $250 Volts,$ having negligible internal resistance. Two resistances $R_{1} and R_{2}$ are $4000 Ω$ and $6000 Ω$ respectively, then :